### Created: May 12^{th}, 2018 | Hotfix 22.19.1

### Last Updated: March 7^{th}, 2020 | Hotfix 27.2.2

- Updated MatLab code a bit:
- Can now run 10 million samples in a more reasonable amount of time (before, going over 1 mil took a noticeably long time). Though, depending on the number of probabilities entered, it can still take upwards of 45 minutes for 10 million samples while 1 mil may take closer to 4 minutes.
- Added Spline and Cubic interpolation to help visualize the curve and get more accurate integral calculations between data points. This also greatly reduces confidence interval sizes.
- Plots have been revamped to display more information (median, mode, and confidence intervals) and use easier to see gradient colors.
- Plots are now on a log scale in the x-axis. The data distribution is a log-normal curve, so using a log scale is more aesthetically pleasing and logical.

Hello and welcome to my discussion and explanation behind the mysterious Warframe run numbers! It's come to my attention that there are some questions, and general disagreements, behind these numbers so I made this blog to hopefully satisfy all your mathematical concerns.

I was actually supposed to do this a long time ago, but quite honestly I simply got busy and kept forgetting, even when reminded :P Sorry for any inconvenience this may have caused, but I hope you can appreciate my explanation popping up out of the blue!

**DEFINITIONS**

First, before we get to the nitty-gritty stuff we should define some key terms to get rid of any confusion and dismiss some overly-literal misinterpretations.

1. **"at least once"**: This sounds pretty straight forward but I want to make sure the concept of receiving each component at least once is crystal clear. Let's take Excalibur's page as an example:

Drop | Chance | Expected | Nearly Guaranteed |
---|---|---|---|

Chassis Blueprint | 38.72% | 6 – 7 | 27 ± 9 |

Neuroptics Blueprint | 38.72% | ||

Systems Blueprint | 22.56% |

In essence, what these values are saying is that in a *total* of 6-7 runs you are expected to receive at least one Neuroptics, at least one Chassis, and at least once Systems component. This is *not* saying that 6-7 runs are needed for *each* individual component; i.e not 18-21 runs in total for all three components.

2. **Expected**: This means the *average* number of runs across the playerbase. In other words, the statistically average player will receive all Warframe components in question at least once in $ x $ number of runs.

3. **Nearly Guaranteed**: This term is probably the one that needs the most clarification. What this describes is the number of runs a player would need to give a higher *probability* of receiving a component at least once. Notice the emphasis and use of "probability" as opposed to "drop chance". Increasing the number of runs **does not** increase the drop chance of a component, but rather, it increases the probability for a component to have dropped already.

The easiest analogy to this is thinking about flipping a coin. A coin always has a 50% chance to either land on heads or tails when flipping it, and much like drop chances, this 50% chance does not change no matter how many times the coin is flipped. Each flip will always have a 50% chance to be either heads or tails. However, consider the question "*If I flip a coin 2 times, what is the probability that it lands on heads at least once*". The chance per flip may be 50% for heads, but the probability that it lands on heads at least once within those two throws *is not* 50%, infact it would be 75%.

If you flip the coin thrice the probability for it to have landed on heads at least once within that time now goes up to 87.5%. Four times will result in a probability of 93.75%. 25 times will result in a probability of 99.99999702% (one might say it's *nearly guaranteed* at that point). Remember, the coin itself is not changing, each flip still has a 50% chance to be heads, but you have increasingly larger probabilities to have gotten heads each flip.

With this in mind, the purpose of "Nearly Guaranteed" is to tell you the number of runs a player needs for a 99% - 99.99% certainty of having received each component in question *at least once*. More specifically, in Excalibur's example above, 27 runs are needed for a 99.9% probability, 18 runs are needed for a 99% probability, and 36 runs are needed for a 99.99% probability.

**MATH**

Okay, now that the boring stuff is out of the way, we can get more into the actual math and statistics of these run stats. For this we will consider three different types of frames.

## Type 1

Type 1's are the frames that drop from a single source; e.g each Excal component drops from Lieutenant Lech Kril on War, Mars. To calculate our numbers we will first find the average chance for each part.

For the following example:

Drop chance for Neuroptics = $ \textbf{P}_{\textbf{1}} $

Drop chance for Chassis = $ \textbf{P}_{\textbf{2}} $

Drop chance for Systems = $ \textbf{P}_{\textbf{3}} $

You start out with zero Excal components.

The chance of you receiving a component you do not already own when killing Kril is

- $ { \textbf{Average Chance (AC)} = \left[ \frac{\textbf{1}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}+\textbf{P}_{\textbf{3}}}\right ]^{-1} = \left[ \frac{\textbf{1}}{\textbf{38.72$%$}+\textbf{38.72$%$}+\textbf{22.56$%$}}\right ]^{-1} \\ = \textbf{100$%$} } $

You now have one Excal component, two remain.

The chance of you receiving a component you do not already own is

- $ { \textbf{AC} = \left[ \frac{\textbf{P}_{\textbf{1}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}+\textbf{P}_{\textbf{3}}}\times\frac{\textbf{1}}{\textbf{P}_{\textbf{2}}+\textbf{P}_{\textbf{3}}} ~ + ~ \frac{\textbf{P}_{\textbf{2}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}+\textbf{P}_{\textbf{3}}}\times\frac{\textbf{1}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{3}}} ~ + ~ \frac{\textbf{P}_{\textbf{3}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}+\textbf{P}_{\textbf{3}}}\times\frac{\textbf{1}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}}\right ]^{-1} \\ = \textbf{64.31$%$} } $

You now have two Excal components, one remains.

The chance of you receiving a component you do not already own is

- $ { \textbf{AC} = \left[ \frac{\textbf{P}_{\textbf{1}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}+\textbf{P}_{\textbf{3}}}\times\frac{\textbf{P}_{\textbf{2}}}{\textbf{P}_{\textbf{2}}+\textbf{P}_{\textbf{3}}}\times\frac{\textbf{1}}{\textbf{P}_{\textbf{3}}} ~ + ~ \frac{\textbf{P}_{\textbf{1}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}+\textbf{P}_{\textbf{3}}}\times\frac{\textbf{P}_{\textbf{3}}}{\textbf{P}_{\textbf{2}}+\textbf{P}_{\textbf{3}}}\times\frac{\textbf{1}}{\textbf{P}_{\textbf{2}}} ~ + ~ \\ \frac{\textbf{P}_{\textbf{2}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}+\textbf{P}_{\textbf{3}}}\times\frac{\textbf{P}_{\textbf{1}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{3}}}\times\frac{\textbf{1}}{\textbf{P}_{\textbf{3}}} ~ + ~ \frac{\textbf{P}_{\textbf{2}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}+\textbf{P}_{\textbf{3}}}\times\frac{\textbf{P}_{\textbf{3}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{3}}}\times\frac{\textbf{1}}{\textbf{P}_{\textbf{1}}} ~ + ~ \\ \frac{\textbf{P}_{\textbf{3}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}+\textbf{P}_{\textbf{3}}}\times\frac{\textbf{P}_{\textbf{1}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}}\times\frac{\textbf{1}}{\textbf{P}_{\textbf{2}}} ~ + ~ \frac{\textbf{P}_{\textbf{3}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}+\textbf{P}_{\textbf{3}}}\times\frac{\textbf{P}_{\textbf{2}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}}\times\frac{\textbf{1}}{\textbf{P}_{\textbf{1}}}\right ]^{-1} \\ = \textbf{28.67$%$} } $

Now that we have the average chances we can simply calculate the Expected number of runs with basic math.

- $ { \textbf{Expected} = \frac{\textbf{1}}{\textbf{100$%$}} ~ + ~ \frac{\textbf{1}}{\textbf{64.31$%$}} ~ + ~ \frac{\textbf{1}}{\textbf{28.67$%$}}\\ = \textbf{6.0429 (6-7)} } $

To calculate the Nearly Guaranteed (NG) number of runs we need some numerical methods. The equation to calculate "at least once" probability from $ n $ percent chances after $ x $ number of runs is:

- $ { \textbf{1} - (\textbf{1} - \textbf{P1})^{x} - (\textbf{1} - \textbf{P2})^{x} - ... - (\textbf{1} - \textbf{PN})^{x} = \textbf{Desired Probability (DP)} } $

But since we want to know $ x $ then we'll need to use numerical methods to solve for it, since we can't rearrange this transcendental equation do so. For our purposes Inverse Quadratic Interpolation will work just fine, so we'll use that.

If you're unfamiliar with IQI it's basically a method for finding the the x-axis value where the root of a function exists. For example, say our equation was much simpler:

- $ { \textbf{2} * x = \textbf{4} } $

If we did not want to solve for $ x $ algebraically we could use IQI to find it for us. First, we must rearrange the equation such that it is equal to zero, like so:

- $ { \textbf{2} * x - \textbf{4} = 0 } $

Next, we will use the IQI function to find at what value of $ x $ the equation is satisfied (in MatLab this function is called `fzero()`

).

x = fzero( @(x) 2 * x - 4 , 0); %The ", 0);" at the end is the x value the function will start at when looking for the answer, it could be any number though.

Almost immediately `fzero()`

will spit out the answer 2

Similarly we will use this method to find the $ x $ value in our previous equation.

First, we rearrange:

- $ { \textbf{1} - \textbf{DP} - (\textbf{1} - \textbf{P1})^{x} - (\textbf{1} - \textbf{P2})^{x} - ... - (\textbf{1} - \textbf{PN})^{x} = 0 } $

Next, we use IQI to find our $ x $:

x = fzero( @(x) 1 - DP - (1 - P1)^(x) - (1 - P2)^(x) - ... - (1 - PN)^(x) , 0);

We'll do this three times, each time with a different DP value of either 99%, 99.9%, and 99.99%. For our Excalibur example above, the results will be:

x = fzero( @(x) 1 - 0.999 - 2*(1 - 0.3872)^(x) - (1 - 0.2256)^(x) , 0);

- $ { \textbf{NG} = \textbf{27}\pm\textbf{9} } $

## Type 2

Type 2's are the frames that have components that drop from separate locations, e.g Harrow's Chassis drops from Fissure Corrupted enemies, his Neuroptics drop from the Pago, Kuva Fortress Spy mission, and his Systems drop from Defection missions. Because these parts are not in the same drop pool as each other we only need to worry about the probabilities of the individual components, which makes these calculations fast and easy.

For the following example:

Drop chance for Neuroptics = 11.28%

Drop chance for Chassis = 3.00%

Drop chance for Systems in Tier I = 7.52%

Drop chance for Systems in Tier II & III = 11.28%

- $ { \textbf{Neuroptics Expected} = \frac{\textbf{1}}{\textbf{11.28$%$}} = \textbf{8.87 (8-9)} } $
- $ { \textbf{Chassis Expected} = \frac{\textbf{1}}{\textbf{3.00$%$}} = \textbf{33.}\overline{\textbf{33}}~\textbf{(33-34)} } $
- $ { \textbf{T1 Systems Expected} = \frac{\textbf{1}}{\textbf{7.52$%$}} = \textbf{13.3 (13-14)} } $
- $ { \textbf{T2 & T3 Systems Expected} = \frac{\textbf{1}}{\textbf{11.28$%$}} = \textbf{8.87 (8-9)} } $

To calculate the Nearly Guaranteed (NG) number of runs we use the same equation as before, however because each part is in its own separate location the equation is no longer trancedental, i.e. $ x $ can be solve for algebraically:

- $ { \textbf{1} - (\textbf{1} - \textbf{11.28$%$})^{x} = \textbf{99.9$%$} } $

- $ { (\textbf{1} - \textbf{11.28$%$})^{x} = (\textbf{1} - \textbf{99.9$%$}) } $

- $ { \frac{\ln(\textbf{1} - \textbf{99.9$%$})}{\ln(\textbf{1} - \textbf{11.28$%$})} = x } $

- $ { \log_{(\textbf{1} - \textbf{11.28$%$})} (\textbf{1} - \textbf{99.99$%$}) = \textbf{58} } $

- $ { \textbf{NG (Neuroptics)} = \textbf{58}\pm\textbf{20} } $

## Type 3

Finally, the third type is really just Equinox. Equinox drops from a single location like type 1 frames, but unlike any other frame, she has eight components to farm for rather than the traditional three. Ideally we'd use the same method as in the type 1 explanation, with five more equations to account for all the parts, however I'll explain why that turns out to be ridiculous.

The equations used for the type one frames are characterized by a pattern of increasing terms. That is, the first equation had just 1 term ([u]), the second had 3 terms ([u + v + w]), and the third had 6 terms ([u + v + w + x + y + z]). To better gauge the pattern we can look at the equations for two components:

- $ {\left[\frac{\textbf{1}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}}\right]^{-1}} $ ([u]), 1 term
- $ { \left[ \frac{\textbf{P}_{\textbf{1}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}}\times\frac{\textbf{1}}{\textbf{P}_{\textbf{2}}} ~ + ~ \frac{\textbf{P}_{\textbf{2}}}{\textbf{P}_{\textbf{1}}+\textbf{P}_{\textbf{2}}}\times\frac{\textbf{1}}{\textbf{P}_{\textbf{1}}}\right ]^{-1} } $ ([u + v]), 2 terms

And for four components the pattern would be:

- [u] (1 term)
- [u + v + w + x] (4 terms)
- [u + v + w + x + y + z + u` + v` + w` + x` + y` + z`] (12 terms)
- [u + v + w + x + y + z + u` + v` + w` + x` + y` + z` + u`` + v`` + w`` + x`` + y`` + z`` + u``` + v``` + w``` + x``` + y``` + z```] (24 terms)

Infact, the number of terms for any number of components ($ n $) in each equation can be found using this, where $ r $ is the number of components already owned:

- $ \frac{n!}{(n - r)!} \text{or} ~ n\textbf{P}r $

What this means is that if we wanted to come up with the 8 equations needed for Equinox, the number of terms in each equation would be:

- 1
- 8
- 56
- 336
- 1,680
- 6,720
- 20,160
- 40,320

So if I wanted to calculate the average chance for just the last component needed, my equation would have 40,320 terms...I'm not doing that. Instead, we can recognize that all 8 of Equinox's component drop chances are relatively close to each other (6 having 12.91% drop chances, the other 2 having 11.28%). Because they're so close to each other we can *estimate* the average chances by assuming each component has a drop chance of 1/8 (12.5%); that they're all evenly distributed. By doing this we can simply use logic to find the average chances:

You start out with zero Equinox components.

The chance of you receiving a component you do not already own when killing Tyl Regor is

- $ { \frac{\textbf{8}}{\textbf{8}} } $

You now have one Equinox component.

The chance of you receiving a component you do not already own is now

- $ { \frac{\textbf{7}}{\textbf{8}} } $

You now have two Equinox components.

The chance of you receiving a component you do not already own is now

- $ { \frac{\textbf{6}}{\textbf{8}} } $

You now have three Equinox components.

The chance of you receiving a component you do not already own is now

- $ { \frac{\textbf{5}}{\textbf{8}} } $

You now have four Equinox components.

The chance of you receiving a component you do not already own is now

- $ { \frac{\textbf{4}}{\textbf{8}} } $

You now have five Equinox components.

The chance of you receiving a component you do not already own is now

- $ { \frac{\textbf{3}}{\textbf{8}} } $

You now have six Equinox components.

The chance of you receiving a component you do not already own is now

- $ { \frac{\textbf{2}}{\textbf{8}} } $

You now have seven Equinox components.

The chance of you receiving the last component you do not already own is

- $ { \frac{\textbf{1}}{\textbf{8}} } $

Now that we have the estimated average chances we can simply calculate the Expected number of runs with basic math like before.

- $ { \textbf{Expected} = \frac{\textbf{8}}{\textbf{8}} ~ + ~ \frac{\textbf{8}}{\textbf{7}} ~ + ~ \frac{\textbf{8}}{\textbf{6}} ~ + ~ \frac{\textbf{8}}{\textbf{5}} ~ + ~ \frac{\textbf{8}}{\textbf{4}} ~ + ~ \frac{\textbf{8}}{\textbf{3}} ~ + ~ \frac{\textbf{8}}{\textbf{2}} ~ + ~ \frac{\textbf{8}}{\textbf{1}}\\ = \textbf{21.7429 (21-22)} } $

To calculate the Nearly Guaranteed (NG) number of runs we use the same methods as in Type 1:

x = fzero( @(x) 1 - 99.9% - 6*(1 - 12.91%)^(x) - 2*(1 - 11.28%)^(x) , 0);

- $ { \textbf{NG} = \textbf{69}\pm\textbf{18} } $

**VALIDITY**

We have the numbers now, but are they actually correct? I cross-checked my work and numbers with some of the methods other people have used; looking at Reddit and /d/ posts claiming the stats are wrong and this or that is the actual correct way, but every time someone made claims like this the comments would always have disagreements with no conclusions. It seemed that no matter where I looked no one could come to a consensus as to what a correct method really was. As such, I just took things into my own hands and created a program to simulate drops.

In short, the program will count the number of "runs" it takes until all parts have been collected, store said count into an array, then repeat the process. Every new iteration of the program repeating these steps represents a different player farming for all the parts. Right now I have the program set to loop through 10,000,000 iterations, each time filling the array to keep track of how many players received all the parts in x number of runs.

For instance, if I were to set the sample size to only 5 instead of 10,000,000 and input 1/3–1/3–1/3 as the drop chances, the array might look like this:

- Runs | Players
- 1 | 0
- 2 | 0
- 3 | 2
- 4 | 0
- 5 | 2
- 6 | 1

What this tells us is that of the 5 players, 2 players got all three parts is just 3 runs, 2 players got all three parts in 5 runs, and 1 player got all three parts in 6 runs.

After the program runs through all 10,000,000 iterations it will plot the data in the array and calculate the *average* number of runs. This, of course, directly corresponds to the Expected number of runs we are calculating with our above methods and equations and will give us a confirmation as to if said methods are the right direction.

For example, recall our Excalibur example and the numbers we calculated:

- $ { \textbf{Expected} = \textbf{6.0429 (6-7)} } $
- $ { \textbf{NG} = \textbf{27}\pm\textbf{9} } $

Using the simulation we are given plots such as this.

*The simulated average (6.3915) is a bit higher than our calculated value (6.0429), but the NG values (28 ± 9) are not too far off from our estimates (27 ± 9).*

We can also do the same thing for Equinox:

- $ { \textbf{Expected} = \textbf{21.7429 (21-22)} } $
- $ { \textbf{NG} = \textbf{69}\pm\textbf{18} } $

*This was simulated with the actual 12.91% - 11.28% values, not the simplified 1/8 distribution we used for our calculations.*

Keep in mind that this is with 10,000,000 samples, for even more accuracy we could increase the sample size to 100,000,000 or even 1,000,000,000. Though quite honestly my computer has a hard enough time running 10,000,000 iterations, so I'll stick to that number or less.

With that said, I believe the simulation and plots give a more than adequate evidence to the validity of our methods above.

**SUGGESTIONS**

Forgetting about the math and technical stuff for a second, what can I do to improve the Expected/NG runs sections of Warframe pages to help clear up confusion in the future?

If you selected Other in any of these polls please let me know in the comments below. I'd love to read and discuss any of your feedback for how to better these sections. Also, *please* let me know if you have questions, comments, or concerns about any of my math or explanations throughout this blog. If there is something fundamental I missed, or you're simply confused on something I failed to fully elaborate on, let me know!

Code for the simulation program can be found below in both .m and .txt format for MatLab and R respectively. If you do not own MatLab I would recommend downloading R as it is free. The only down side is you can't make the plots as pretty as I have them here ¯\_(ツ)_/¯. You can also still try downloading the .m or .txt file and use a converter to change it to your desired language, however, I can not guarantee the reliability nor accuracy of said converters:

Thank you!